Q91 / 180 · KCET 2015 · Mathematics
If the function \( g(x) \) is defined by
\[
g(x)=\frac{\chi^{200}}{200}+\frac{\chi^{199}}{199}+\frac{\chi^{198}}{198}+\ldots \ldots+\frac{\chi^{2}}{2}+\chi+5, \text { then } g^{\prime}(0)=
\]
\[ g(x)=\frac{\chi^{200}}{200}+\frac{\chi^{199}}{199}+\frac{\chi^{198}}{198}+\ldots \ldots+\frac{\chi^{2}}{2}+\chi+5, \text { then } g^{\prime}(0)= \]
- Correct answerA\( 11 \)
- B\( 200 \)
- C\( 100 \)
- D\( 55 \)
Solution
Given that,
\[ \begin{array}{l} g(x)=\frac{\chi^{200}}{200}+\frac{\chi^{199}}{199}+\frac{\chi^{198}}{198}+\cdot s+\frac{\chi^{2}}{2}+\chi+5 \\ \text { So, } g^{\prime}(x)=\frac{200 x^{199}}{200}+\frac{199 x^{198}}{199}+s+\frac{2 x}{2}+1+0 \end{array} \]
At \( x=0 \), we have \( g^{\prime}(0)=1 \)
\[ \begin{array}{l} g(x)=\frac{\chi^{200}}{200}+\frac{\chi^{199}}{199}+\frac{\chi^{198}}{198}+\cdot s+\frac{\chi^{2}}{2}+\chi+5 \\ \text { So, } g^{\prime}(x)=\frac{200 x^{199}}{200}+\frac{199 x^{198}}{199}+s+\frac{2 x}{2}+1+0 \end{array} \]
At \( x=0 \), we have \( g^{\prime}(0)=1 \)
Hence, the correct answer is option: 'A'.